Problem

Source: Australia MO 2024 P7

Tags: geometry



Let $ABCD$ be a square and let $P$ be a point on side $AB$. The point $Q$ lies outside the square such that $\angle ABQ = \angle ADP$ and $\angle AQB = 90^{\circ}$. The point $R$ lies on the side $BC$ such that $\angle BAR = \angle ADQ$. Prove that the lines $AR, CQ$ and $DP$ pass through a common point.